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From: Mikko <mikko.levanto@iki.fi>
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Subject: =?utf-8?Q?Re:_A_simulating_halt_decider_applied_to_the_The_Peter_Linz_Turing_Machine_description_=E2=9F=A8=C4=A4=E2=9F=A9_--_key_details?=
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On 2024-06-02 13:21:56 +0000, olcott said:

> On 6/2/2024 2:42 AM, Mikko wrote:
>> On 2024-06-01 19:26:55 +0000, olcott said:
>> 
>>> On 6/1/2024 1:52 PM, joes wrote:
>>>> Am Sat, 01 Jun 2024 09:37:01 -0500 schrieb olcott:
>>>>> On 6/1/2024 2:52 AM, Mikko wrote:
>>>>>> On 2024-05-31 15:35:18 +0000, olcott said:
>>>> 
>>>>>>> *A quick summary of the reasoning provided below*
>>>>>>> The LHS is behavior that embedded_H is allowed to report on.
>>>>>> There is no restrictions on what embedded_H is allowed to report on.
>>>>> 
>>>>> embedded_H is only allowed to report on the behavior that its finite
>>>>> string Turing Machine Description specifies to a UTM.
>>>>> 
>>>>> embedded_H <is> a UTM except that it stops simulating and reports
>>>>> non-halting as soon as it correctly recognizes a non-halting behavior
>>>>> pattern that is specified by its input.
>>>> "Except". So it is not an UTM.
>>>> 
>>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>> 
>>>>> (a) Ĥ copies its input ⟨Ĥ⟩
>>>>> (b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>> (c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>> (d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
>>>>> (e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>> (f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>> (g) goto (d)
>>>>> 
>>>>> embedded_H is not allowed to be applied to Ĥ ⟨Ĥ⟩ because inputs can
>>>>> only be finite strings and Ĥ is not a finite string. This means
>>>>> that embedded_H is not allowed to report on its own actual behavior.
>>>> I can't read that notation. What is H^ and what does it look like?
>>>> 
>>> 
>>> *Here is the whole Linz proof*
>>> I simplified the Linz notation at the bottom of page 2 of the proof.
>>> https://www.liarparadox.org/Linz_Proof.pdf
>> 
>> You are right, that is a sufficient proof. You may change the presentation
>> but then you must prove that your presentation is equivalent to Linz'.
> 
> The proof was removed until Joe could understand what Linz was saying
> Here is my actual proof.

Linz' proof is Linz' proof, not yours.

> When Ĥ is applied to ⟨Ĥ⟩
> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The above does not say what Linz said and hardly anything else, either.
It is not a sentence. If you remove the second last or the last line
or add some conjunction between them and add a point to the end then
you would have a sentence.

> (a) Ĥ copies its input ⟨Ĥ⟩
> (b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
> (c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
> (d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
> (e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
> (f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
> (g) goto (d)
> 
> Linz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ derives a different result than
> embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩.

Irrelevant as embedded_H is not a part of Linz' proof. The part
of Linz' Ĥ that corresponds to your embedded_H does produce the
same result as Linz H does.

Turns out that you can prove that the result produced by Linz' H is
different from the result produced by linz' H. This is sufficient to
prove that Linz' H does not exist.

> This is because the in the latter case embedded_H must determine that
> ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly stop running
> after 1 to ∞ steps of correct simulation. Thus (as we can all see) 
> embedded_H meets its abort simulation criteria.

Those criteria are not mentioned in Linz' proof and are therefore
irrelevant to it.

-- 
Mikko